Context & Interest.
See this MSE post about a twin-prime related topology. Basically $0$ is a easily seen to be a generic point in this topology. Every generic point is clearly dense as a singleton subset, meaning its closure is the whole space $\Bbb{Z}/p_n\#$. In an approach toward potentially proving the twin prime conjecture, I must show that the interval of residues $I = [1, \dots, p_{n+1}^2 - 2]$ is such that its set of elementwise powers $\mathcal{I} = \{ x^k : x \in I, k \geq 0\}$ is also dense in $(\Bbb{Z}, \tau)$. But because $x = 2,3$ are such that $x^2 \neq 1 \pmod q$ for all $q\geq 5$, we get reduction to the below question. Or in other words the conjecture below is sufficient to prove density of $\mathcal{I}$.
Conjecture. For all sufficiently large $n$ (something like $n \geq 3$); either $\exists k \geq 2$ such that $\gcd(2^k - 1, \frac{p_n\#}{6}) = 1$, or $\exists k \geq 2$ such that $\gcd(3^k - 1, \frac{p_n\#}{6}) = 1$, or both.
Where $p_n\# := p_{n}p_{n-1} \cdots p_1$ is primorial.
Question. How can we prove this conjecture easily (intuitively it looks like an easy problem)?
Attempt. Working each possibility in the either/or individually, take the first one.
Then we want to show that $2^k - 1$ is a solution to:
$$ X(X-2)\cdots (X-(q - 2)) = 0 \pmod {q}, \\ \forall\text{ prime } q \mid p_n\# $$
for some $k \geq 0$. But this seems more hopeless than the original formulation!
Yes, your conjecture is true. In particular, for any $n\ge 3$, then $\exists\, k \geq 3$ such that $\gcd(2^k - 1, \frac{p_n\#}{6}) = 1$, and $\exists\, k \geq 3$ such that $\gcd(3^k - 1, \frac{p_n\#}{6}) = 1$. In particular, this occurs with any prime $k$ where $2k + 1 \gt p_n$. First, for $2^k - 1$, for any prime $p \mid 2^k - 1$, then
$$2^k \equiv 1 \pmod{p}$$
With the multiplicative order, we get
$$m = \operatorname{ord}_{p}(2) \;\;\to\;\; m \mid k$$
Since $k$ is prime, this means that $m = 1$ or $m = k$. As $m = 1$ doesn't work, this means that $m = k$. However, by Euler's theorem, we then get for some positive, even integer $j$ that
$$k \mid \varphi(p) \;\to\; p - 1 \equiv 0 \pmod{k} \;\to\; p = jk + 1 \;\to\; p \ge 2k + 1 \gt p_n$$
As this applies to all prime factors of $2^k - 1$, we then get
$$\gcd\left(2^k - 1, \frac{p_n\#}{6}\right) = 1$$
A similar argument shows that $\gcd(3^k - 1, \frac{p_n\#}{6}) = 1$.
More generally, $k$ can have up to $1$ factor of $2$ (since then the multiplicative order $m$ of any prime $p \mid a^k$ where $a\in\{2,3\}$ can possibly have a factor of $2$ but, since $2^2 - 1 = 3$ and $3^2 - 1 = 8$, then $m$ is not $2$, so it needs to have at least one odd prime factor of $k$) and every odd prime factor $p$ has $2p + 1 \gt p_n$.